$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$

Question

$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ (use Cauchy–Schwarz inequality)

I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?

I need a hint

Answer 1

Hint: Use Cauchy Schwarz in Engelform: $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\geq \frac{(x+y+z)^2}{a+b+c}$$ It is equivalent to $${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2} \geq 0$$ and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2\geq 0$$ if $$a,b,c$$ are positive.

Answer 2

By C-S $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=\frac{\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)(a+b+c)}{a+b+c}\geq$$ $$\geq\frac{\left(\frac{x}{\sqrt{a}}\cdot\sqrt{a}+\frac{y}{\sqrt{b}}\cdot\sqrt{b}+\frac{z}{\sqrt{c}}\cdot\sqrt{c}\right)^2}{a+b+c}=\frac{(x+y+z)^2}{a+b+c}.$$ The equality occurs for $$\left(\frac{x}{\sqrt{a}},\frac{y}{\sqrt{b}},\frac{z}{\sqrt{c}}\right)||(\sqrt{a},\sqrt{b},\sqrt{c})$$ or $$(x,y,z)||(a,b,c).$$ Actually, the last writing is very useful.